## PLEASE~ tell me if i'm doing these correctly

This is the first question for my assignment...I've used a tree diagram to solve this one. Please~~ if anyone can tell me if i'm doing this correctly!!!! APPRECIATED!

Team A and Team B completing, the first who win 2 games wins the series. Team A has probability of 0.6 of winning the game. And Team B has the probability of 0.6 of winning the next 2 games, IF NECESSARY. Assuming the outcomes for different games are independent.

<a> Find the probability that Team A will win the series.
Let A1 = Team A wins the 1st game
P(A1) = 0.6
A2 = Team A wins the 2nd game
P(A2) = 0.4
A3 = Team A wins the 3rd game (if Team A lost the 2nd game)
P(A3) = 0.4

P(A1Ո A2) = P(A1) P(A2)
= 0.24
P(A1Ո A2’ Ո A3) = (0.6)(0.6)(0.4)
= 0.144

P(A1’ Ո A2 Ո A3) = (0.4)(0.4)(0.4)
= 0.064

Team A wins = P(A1Ո A2) + P(A1Ո A2’ Ո A3) + P(A1’ Ո A2 Ո A3)
= 0.448

<b> Given that Team A won the first game, what is the probability that they will go on to win the series?

P(A2 | A1) = 0.4
P (A2’ Ո A3 | A1) = (0.6) (0.4) = 0.24

Therefore the probability that Team A wins the series given that they won the first game
= P(A2 | A1) + P (A2’ Ո A3 | A1) = 0.64

<c> Find the probability that it takes 3 games to decide the series.
P(A1 Ո A2’ Ո A3) = (0.6)(0.6)(0.4) = 0.144

P(A1 Ո A2’ Ո A3’) = (0.6)(0.6)(0.6) = 0.216

P(A1’ Ո A2 Ո A3) = (0.4)(0.4)(0.4) = 0.064

P(A1’ Ո A2 Ո A3’) = (0.4)(0.4)(0.6) = 0.096

Probability that it takes 3 games to decide the series
= P(A1 Ո A2’ Ո A3) + P(A1 Ո A2’ Ո A3’) + P(A1’ Ո A2 Ո A3) + P(A1’ Ո A2 Ո A3’)
= (0.144) + (0.216) + (0.064) + (0.096)
=0.52

Sorry guys...abit long...but...i'm trying to be as clear as possible of what i've done!